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A box contains 11 slips of paper. The slips are labeled with the first 11 prime numbers, with one different prime number on each slip. If 4 slips are drawn at random from the box without replacement, what is the probability that the range of the 4 numbers drawn is 12?
A.
\(\frac{1}{66}\)
B.
\(\frac{1}{30}\)
C.
\(\frac{1}{22}\)
D.
\(\frac{1}{15}\)
E.
\(\frac{1}{11}\)
A.
\(\frac{1}{66}\)
B.
\(\frac{1}{30}\)
C.
\(\frac{1}{22}\)
D.
\(\frac{1}{15}\)
E.
\(\frac{1}{11}\)
07 Aug 2026, 05:15
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Official Solution:
A box contains 11 slips of paper. The slips are labeled with the first 11 prime numbers, with one different prime number on each slip. If 4 slips are drawn at random from the box without replacement, what is the probability that the range of the 4 numbers drawn is 12?
A.
\(\frac{1}{66}\)
B.
\(\frac{1}{30}\)
C.
\(\frac{1}{22}\)
D.
\(\frac{1}{15}\)
E.
\(\frac{1}{11}\)
The range is the positive difference between the greatest chosen number and the smallest chosen number.
The first 11 prime numbers are:
\(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31\)
For the range to be 12, the smallest and greatest chosen numbers must differ by 12.
The possible endpoint pairs are:
• 5 and 17
• 7 and 19
• 11 and 23
• 17 and 29
• 19 and 31
For each pair, the two endpoints are fixed, and there is only 1 way to choose that pair. The remaining 2 numbers must be chosen from the primes strictly between those endpoints.
For 5 and 17: choose 2 from 7, 11, 13, giving \(C(3,2) = 3\) ways.
For 7 and 19: choose 2 from 11, 13, 17, giving \(C(3,2) = 3\) ways.
For 11 and 23: choose 2 from 13, 17, 19, giving \(C(3,2) = 3\) ways.
For 17 and 29: choose 2 from 19, 23, giving \(C(2,2) = 1\) way.
For 19 and 31: choose 2 from 23, 29, giving \(C(2,2) = 1\) way.
So the number of favorable selections is:
\(3 + 3 + 3 + 1 + 1 = 11\)
The total number of ways to choose 4 slips from 11 is:
\(C(11,4) = 330\)
Therefore, the probability is:
\(\frac{11}{330} = \frac{1}{30}\)
Answer: B
A box contains 11 slips of paper. The slips are labeled with the first 11 prime numbers, with one different prime number on each slip. If 4 slips are drawn at random from the box without replacement, what is the probability that the range of the 4 numbers drawn is 12?
A.
\(\frac{1}{66}\)
B.
\(\frac{1}{30}\)
C.
\(\frac{1}{22}\)
D.
\(\frac{1}{15}\)
E.
\(\frac{1}{11}\)
The range is the positive difference between the greatest chosen number and the smallest chosen number.
The first 11 prime numbers are:
\(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31\)
For the range to be 12, the smallest and greatest chosen numbers must differ by 12.
The possible endpoint pairs are:
• 5 and 17
• 7 and 19
• 11 and 23
• 17 and 29
• 19 and 31
For each pair, the two endpoints are fixed, and there is only 1 way to choose that pair. The remaining 2 numbers must be chosen from the primes strictly between those endpoints.
For 5 and 17: choose 2 from 7, 11, 13, giving \(C(3,2) = 3\) ways.
For 7 and 19: choose 2 from 11, 13, 17, giving \(C(3,2) = 3\) ways.
For 11 and 23: choose 2 from 13, 17, 19, giving \(C(3,2) = 3\) ways.
For 17 and 29: choose 2 from 19, 23, giving \(C(2,2) = 1\) way.
For 19 and 31: choose 2 from 23, 29, giving \(C(2,2) = 1\) way.
So the number of favorable selections is:
\(3 + 3 + 3 + 1 + 1 = 11\)
The total number of ways to choose 4 slips from 11 is:
\(C(11,4) = 330\)
Therefore, the probability is:
\(\frac{11}{330} = \frac{1}{30}\)
Answer: B
